Class XII NCERT Maths Chapter 2- Inverse …...Class XII – NCERT – Maths Chapter 2- Inverse...

Class XII – NCERT – Maths Chapter 2- Inverse trigonometric functions

2.Inverse trigonometric functions

Exercise 2.1

Question 1:

Find the principal value of 1 1sin

2

Solution 1:

Let 1 1sin ,

2y

Then 1

sin2

y

sin sin .6 6

We know that the range of the principal value branch of 1sin is ,

2 2

1sin ,

6 2and

Therefore, the principal value of 1 1sin .

2 6is

Question 2:

Find the principal value of 1 3cos

2

Solution 2:

Let 1 3cos

2y

. Then

3cos cos

2 6y

We know that the range of the principal value branch of 1cos is 0,

3cos

6 2and

.

Therefore, the principal value of 1 3cos

2 6is

.

Question 3:

Find the principal value of 1cosec (2)

Solution 3:



Let 1cosec (2) y .

Then, cosec y = 2= cosec .6

We know that the range of the principal value branch of -1cosec is , 0 .

2 2

Therefore, the principal value of -1cosec 2 is .6

Question 4:

Find the principal value of 1tan 3

Solution 4:

Let 1tan 3 y

Then, tan 3 tan tan .3 3

y

We know that the range of the principal value branch of 1tan is

2 2

tan 3.3

and is

Therefore, known that the principal value of 1tan 3 is 3

.

Question 5:

Find the principal value of 1 1cos

2

Solution 5:

1 1cos .

2Let y

Then, 1 2

cos cos cos cos .2 3 3 3

y

We know that the range of the principal value branch of 1cos is 0,

2 1cos .

3 2and



Therefore, the principal value of 1 1cos

2

is 2

3

.

Question 6:

Find the principal value of 1tan 1

Solution 6:

Let 1tan 1 = y.

Then, tan 1 tan tan .4 4

y

We know that the range of the principal value branch of 1tan is ,

2 2

tan 1.4

and

Therefore, the principal value of 1tan 1 is 4

.

Question 7:

Find the principal value of 1 2

sec3

Solution 7:

Let 1 2

sec3

y

. Then,

2sec sec .

63y

We know that the range of the principal value branch of 1sec is 0,

2

2sec .

6 3and

Therefore, the principal value of 1 2

sec .63

is

https://www.ncertbooks.guru/cbse-ncert-solutions-pdf/



Question 8:

Find the principal value of 1cot 3

Solution 8:

Let 1cot 3 y . Then cot 3 cot .6

y

We know that the range of the principal value branch of 1cot is (0, )

and cot 3.6

Therefore, the principal value of 1cot 3 is

6

.

Question 9:

Find the principal value of 1 1

cos2

Solution 9:

Let 1 1

cos2

y

.

Then 1 3

cos cos cos cos .4 4 42

y

We know that the range of the principal value branch of 1cos is [0, ]

and 3 1

cos4 2

.

Therefore, the principal value of 1 1

cos2

is 3

4

.

Question 10:

Find the principal value of -1cosec 2

Solution 10:



-1Let cosec 2 .y Then, cos 2ec y cos cos .4 4

ec ec

We know that the range of the principal value branch of -1cosec is

, {0}2 2

and cosec 24

.

Therefore, the principal value of -1cosec 2 is 4

.

Question 11:

Find the value of 1 1 11 1tan 1 cos sin

2 2

Solution 11:

Let 1tan 1 .x

Then, tan =1= tan .4

x

1tan 14

Let 1 1cos .

2y

Then, 1 2

cos cos cos cos .2 3 3 3

y

-1 1 2 cos

2 3

Let 1 1sin .

2z

Then, 1

sin sin sin .2 6 6

z

1 1sin

2 6

1 1 11 1tan 1 cos sin

2 2



2

4 3 6

3 8 2 9 3

12 12 4

Question 12:

Find the value of 1 11 1cos 2sin

2 2

Solution 12:

1 1cos .

2Let x

1Then,cos cos .

2 3x

1 1cos

2 3

1 1sin .

2Let y

Then, 1

sin sin .2 6

y

1

1 1

1sin

2 6

1 1 2cos 2sin 2

2 2 3 6 3 3 3

Question 13:

If 1sin , thenx y

(A) 0 y (B) 2 2

y

(C) 0 y (D) 2 2

y

Solution 13:



It is given that 1sin x y .

We know that the range of the principal value branch of 1sin

, .2 2

is

Therefore, .2 2

y

Answer choice (B) is correct.

Question 14:

1 1tan 3 sec 2 is equal to

(A) (B) / 3

(C) / 3 (D) 2 / 3

Solution 14:

Let 1tan 3 .x .

Then, tan 3 tan3

x

We know that the range of the principal value branch of 1tan is , .

2 2

1tan 33

Let 1sec 2 .y

Then, 2

sec 2 sec sec sec .3 3 3

y

We know that the range of the principal value branch of 1sec is 0, .2

1 2sec 2

3

Thus,

1 1tan 3 sec 2

2

3 3 3



Exercise 2.2

Question 1:

Prove 1 1 3 1 13sin sin 3 4 , ,

2 2x x x x

Solution 1:

To Prove 1 1 3 1 13sin sin 3 4 , ,

2 2x x x where x

Let 1sin . ,sin .x Then x

We have,

R.H.S

1 3 1 3sin 3 4 sin 3sin 4sinx x

1

1

sin sin 3

3

3sin . .x L H S

Question 2:

Prove 1 1 3 13cos cos 4 3 , ,1

2x x x x

Solution 2:

To Prove 1 1 3 13cos cos 4 3 , ,1

2x x x x

Let 1cos . Then, cosx x

We have

1 3

. .

cos 4 3

R H S

x x



1 3

1

1

cos 4cos 3cos

cos cos3

3

3cos . .x L H S

Question 3:

Prove 1 1 12 7 1

tan tan tan11 24 2

Solution 3:

To prove: 1 1 12 7 1

tan tan tan11 24 2

1 1

1 1 1 1

1

1

1 1

. . .

2 7tan tan

11 24

2 7

11 24tan tan tan tan2 7 1

1 .11 24

11 24tan

11 24 14

11 24

48 77tan

264 14

125 1tan tan R.H.S.

250 2

L H S

x yx y

xy



Question 4:

Prove 1 1 11 1 31

2 tan tan tan2 7 17

Solution 4:

To prove: 1 1 11 1 31

2 tan tan tan2 7 17

-1 11 1L.H.S = 2tan tan

2 7

1 1 1 1

2 2

12.

1 22tan tan 2 tan tan7 11

12

xx

x

1 11 1tan tan

3 7

4

1 14 1tan tan

3 7

1 1 1 1

4 1


1 .3 7

x yx y

xy

1

28 3

21tan21 4

21

1 31tan R.H.S.

17

Question 5:

Write the function in the simplest form: 2

1 1 1tan , 0

xx

x

Solution 5:

21 1 1

tanx

x

Put 1tan tanx x



2 21 11 1 1 tan

tan tantan

x

x

1 1sec 1 1 costan tan

tan sin

2

1

2sin2

tan

2sin cos2 2

1 11tan tan tan

2 2 2x

Question 6:


1

1

1tan , 1

xx

Solution 6:

2

1

1

1tan , 1

xx

Put 1cosec cosx ec x

1

2

1

2

1tan

1

1tan

cos 1

x

ec

1

1

1tan

cot

tan tan

1

1 1 1

cos

sec ,cos sec2 2

ec x

x As ec x x



Question 7:

Write the function in the simplest form: 1 1 costan ,

1 cos

xx

x

Solution 7:

1 1 costan ,

1 cos

xx

x

1

2

1

2

1 costan

1 cos

2sin2tan

2cos2

x

x

x

x

1

1

sin2tan

cos2

tan tan2

x

x

x

2

x

Question 8:

Write the function in the simplest form: 1 cos sintan ,0

cos sin

x xx

x x

Solution 8:

1 cos sintan

cos sin

x x

x x

1

sin1

costan

sin1

cos

x

x

x

x



1 1 tantan

1 tan

x

x

1 1 1 1 1tan 1 tan tan tan tan tan1

x yx x y

xy

=4

x

Question 9:


2 2tan ,

xx a

a x

Solution 9:

1

2 2tan

x

a x

Let, 1sin sin sinx x

x aa a

1

2 2

1

2 2 2

tan

sintan

sin

x

a x

a

a a

1

2

1

sintan

1 sin

sintan

cos

a

a

a

a

1 1tan tan sinx

a

Question 10:

Write the function in the simplest form: 2 3

1

3 2

3tan , 0;

3 3 3

a x x a aa x

a ax

Solution 10:



Consider, 2 3

1

3 2

3tan

3

a x x

a ax

Let

1tan tan tan

x xx a

a a

2 31

3 2

2 3 31

3 2 2

3tan

3

3 tan tantan

3 tan

a x x

a ax

a a a

a a a

3 3 31

3 3 2

3 tan tantan

3 tan

a a

a a

1tan tan3

3

13tanx

a

Question 11:

Find the value of 1 1 1

tan 2cos 2sin2

Solution 11:

Let 1 1

sin .2

x

1,sin sin .

2 6Then x

1 1sin

2 6

1 1

1

1tan 2cos 2sin

2

tan 2cos 26



1

1

tan 2cos3

1tan 2

2

1tan 14

Question 12:

Find the value of 1 1cot tan cota a

Solution 12:

1 1cot tan cota a

1 1cot tan cot2 2

x x

0

Question 13:

Find the value of 2

1 1

2 2

1 2 1tan sin cos , 1, 0

2 1 1

x yx y

x y

and 1xy

Solution 13:

Let tan .x 1, tan .Then x

1 1

2 2

1

1

2 2 tansin sin

1 1 tan

sin sin 2

2

2 tan

x

x

x

Let 1tan . , tan .y Then y



21

2

21

2

1

1

1cos

1

1 tancos

1 tan

cos cos 2

2 2 tan

y

y

y

21 1

2 2

1 2 1tan sin cos

2 1 1

x y

x y

1 11tan 2 tan 2 tan

2x y

1 1 1, tan tan tan1

x yAs x y

xy

1tan tan1

x y

xy

1

x y

xy

Question 14:

If 1 11sin sin cos 1

5x

, then find the value of x.

Solution 14:

1 11sin sin cos 1

5x

1 1 1 1

1 1

1 1

1 1sin sin cos cos cos sin sin cos 1

5 5

sin sin cos cos sin

1 1cos sin sin cos 1

5 5

1cos sin sin cos 1 ...(1)

5 5

x x

A B A B A B

x x

xx

Now, let 1 1sin

5y

Then,



1

2

1

1sin

5

1sin

5

1 2 6cos 1

5 5

2 6cos

5

y

y

y

y

1 11 2 6sin cos ...(2)

5 5

Let 1cos .x z

Then, cos z x

2

1 2

sin 1

sin 1

z x

z x

.

1 1 2cos sin 1 ...(3)x x

From (1), (2) and (3) we have:

1 1 2

2

2

2

2 6cos cos sin sin 1 1

5 5

2 61 1

5 5

2 6 1 5

2 6 1 5

xx

xx

x x

x x

On squaring both sides, we get:



22 2

2 2

2 2

2

2

2 6 1 25 10

4 6 1 25 10

24 24 25 10

25 10 1 0

5 1 0

5 1 0

1

5

x x x

x x x

x x x

x x

x

x

x

Hence, the value of x is 1

5.

Question 15:

If 1 11 1

tan tan ,2 2 4

x x

x x

then find the value of x.

Solution 15:

1 11 1tan tan

2 2 4

x x

x x

1

1 1

2 2tan1 1 4

12 2

x x

x xx x

x x

1 1 1, tan tan tan

1

x yAs x y

xy

1

1 2 1 2tan

2 2 1 ( 1) 4

x x x x

x x x x

2 21

2 2

2 2tan

4 1 4

x x x x

x x

21 2 4

tan3 4

x

21 4 2

tan tan tan3 4

x

24 21

3

x



24 2 3x 22 4 3 1x

1

2x

Hence, the value of x is 1

2 .

Question 16:

Find the values of 1 2sin sin

3

Solution 16:

Consider, 1 2sin sin

3

We know that 1sin sin x x

If ,2 2

x

, which is the principal value branch of 1sin .x

.

Here, 2

,3 2 2

Now, 1 2sin sin

3

can be written as:

1

1

1

2sin sin

3

2sin sin

3

sin sin , ,3 3 2 2

where

1 12sin sin sin sin

3 3 3



Question 17:

Find the values of 1 3tan tan

4

Solution 17:

Consider, 1 3tan tan

4

We know that 1tan tan x x

If ,2 2

x

, which is the principal value branch of 1tan .x

Here, 3

, .4 2 2

Now, 1 3tan tan

4

can be written as:

1 1 13 3tan tan tan tan tan tan

4 4 4

1 1tan tan tan tan4 4

where ,4 2 2

1 13tan tan tan tan

4 4 4

Question 18:

Find the values of 1 13 3tan sin cot

5 2

Solution 18:

Let 1 3

sin5

x .

Then ,

2

3sin

5

4cos 1 sin

5

5sec

4

x

x x

x



2 25 3tan sec 1 1

16 4x x

1 3tan

4x

1 13 3sin tan ...( )

5 4i

Now, 1 13 2

cot tan ...( )2 3

ii 1 11, tan cot x

x

Therefore, 1 13 3tan sin cot

5 2

1 13 2tan tan tan

4 3

[Using (i) and (ii)]

1

3 2

4 3tan tan3 2

14 3

1 1 1, tan tan tan

1

x yAs x y

xy

1 9 8tan tan

12 6

1 17 17tan tan

6 6

Question 19:

Find the values of 1 7cos cos

6

is equal to

7( )

6A

5( )

6B

( )3

C

( )6

D

Solution 19:

We know that 1cos cos x x if 0,x , which is the principal value branch of 1cos x

.

Here, 7

0, .6



Now, 1 7cos cos

6

can be written as :

1 1 17 7 7cos cos cos cos cos cos 2

6 6 6

cos 2 cosx x

1 17 5 5cos cos cos cos

6 6 6

The correct answer is B.

Question 20:

Find the values of 1 1

sin sin3 2

is equal to

(A) 1

2 (B)

1

3

(C) 1

4

(D) 1

Solution 20:

Let 1 1sin

2x

.

Then , 1

sin sin sin .2 6 6

x

We know that the range of the principal value branch of 1sin , .2 2

is

1 1sin

2 6

1 1 3sin sin sin sin sin 1

3 2 3 6 6 2

The correct answer is D.




Miscellaneous Exercise

Question 1:

Find the value of 1 13cos cos

6

Solution 1:

We Know that 1cos cos x x if 0,x , which is the principal value branch of 1cos x

.

Here, 13

0, .6

Now, 1 13cos cos

6

can be written as :

1

1

1

13cos cos

6

cos cos 26

cos cos 0, .6 6

where

1 113cos cos cos cos

6 6 6

Question 2:

Find the value of 1 7tan tan

6

Solution 2:

We know that 1tan tan x x if ,

2 2x

, which is the principal value branch of 1tan .x

Here, 7

, .6 2 2

Now, 1 7tan tan

6

can be written as:



1

1

7tan tan

6

5tan tan 2

6

tan 2 tanx x

1

1

5tan tan tan 2 tan

6

5tan tan

6

x x

1tan tan ,6

where ,

6 2 2

1 17tan tan tan tan

6 6 6

Question 3:

Prove 1 13 24

2sin tan5 7

Solution 3:

Let 1 3

sin .5

x Then 3

sin .5

x

23 4

cos 15 5

x

3tan

4x

1 1 13 3 3tan sin tan

4 5 4x

Now, we have:

L.H.S

1 13 3

2sin 2 tan5 4

1

2

32

4tan3

14

1 1

2

22 tan tan

1

xx

x



1 1

33 162tan tan

16 9 2 7

16

1 24tan . . .

7R H S

Question 4:

Prove 1 1 18 3 77

sin sin tan17 5 36

Solution 4:

1 8sin .

17x

Then,

28 8 225 15

sin cos 117 17 289 17.

x x

18 8tan tan

15 15x x

1 18 8sin tan

17 15

…..(1)

Now, let 1 3

sin5

y

Then, 2

3 3 16 4sin cos 1 .

5 5 25 5y y

13 3tan tan

5 4y y

1 13 3sin tan

5 4

……(2)

Now, we have:

L.H.S.

1 18 3

sin sin17 5



Using (1) and (2), we get

= 1 18 3

tan tan15 4

1 1 1 1

8 3


115 4

x yx y

xy

1 32 45tan

60 24

1 77tan . . .

36R H S

Question 5:

Prove 1 1 14 12 33

cos cos cos5 13 65

Solution 5:

Let 1 4cos

5x

Then, 2

4 4 3cos sin 1

5 5 5x x

13 3tan tan

4 4x x

1 14 3cos tan

5 4

….(1)

Now, let 1 12

cos .13

y

Then, 12 5

cos sin .13 13

y y

15 5tan tan

12 12y y

1 112 5cos tan

13 12

….(2)

Let 1 33

cos65

z



Then, 33 56

cos sin65 65

z z

156 56tan tan

33 33z z

1 133 56cos tan

65 33

...(3)

Now,

1 1

. . .

4 12cos cos

5 13

L H S


1 13 5tan tan

4 12

1 1 1 1

3 5


14 12

x yx y

xy

1 36 20tan

48 15

1 56tan

33

1 56cos

33

. . .R H S

Question 6:

Prove 1 1 112 3 56

cos sin sin13 5 65

Solution 6:

Let 1 3sin .

5x Then,

23 3 16 4

sin cos 1 .5 5 25 5

x x



13 3tan tan

4 4x x

1 13 3sin tan ...(1)

5 4

Now, let 1 12cos

13y

12 5,cos sin .

13 13Then y y

15 5tan tan

12 12y y

1 112 5cos tan ...(2)

13 12

Let 1 56sin .

65z

Then, 56 33

sin cos .65 65

z z

156 56tan tan

33 33z z

1 156 56sin tan ...(3)

65 33

Now, we have

L.H.S

1 112 3cos sin

13 5

1 15 3tan tan

12 4




-1 1 1 1

5 3

12 4= tan tan tan tan5 3 1

1 .12 4

x yx y

xy

= 1 20 36tan

48 15

= 1 56tan

33

1 56sin [Using (3)]

65

. .R H S

Question 7:

Prove 1 1 163 5 3tan sin cos

16 13 5

Solution 7:

Let 1 5sin

13x

Then, 5 12

sin cos .13 13

x x

15 5tan tan

12 12x x

_1 15 5sin tan

13 12

…..(1)

Let 1 3

cos .5

y

Then, 3 4

cos sin .5 5

y y

14 4, tan tan

3 3Thus y y

1 13 4cos tan

5 3

….(2)

Using (1) and (2), we have

R.H.S.



1 15 3sin cos

12 5

= 1 15 4tan tan

12 3

1 1 1 1

5 4


112 3

x yx y

xy

1 15 48tan

36 20

1 63tan

16

L.H.S .

Question 8:

Prove 1 1 1 11 1 1 1tan tan tan tan

5 7 3 8 4

Solution 8:

1 1 1 1

L.H.S

1 1 1 1tan tan tan tan

5 7 3 8

1 1

1 1 1 1

5 7 3 8tan tan1 1 1 1

1 15 7 3 8

1 1 1tan tan tan

1

x yx y

xy

1 17 5 8 3tan tan

35 1 24 1

1 112 11tan tan

34 23

1 16 11tan tan

17 23

1

6 11

17 23tan6 11

117 23



1 1325tan tan 1

325

R.H.S.4

Question 9:

Prove 1 11 1tan cos , 0.1

2 1

xx x

x

Solution 9:

Let 2tan .x

Then 1tan tan .x x

2

2

1 1 tancos 2

1 1 tan

x

x

Now, we have:

1

1

-1

. .

1 1cos

2 1

1cos cos 2

2

12 = =tan . . .

2

R H S

x

x

x L H S

Question 10:

Prove 1 1 sin 1 sincot , 0,

2 41 sin 1 sin

x x xx

x x

Solution 10:



Consider 1 sin 1 sin

1 sin 1 sin

x x

x x

2

2

1 sin 1 sin

1 sin 1 sin

x x

x x

(by rationalizing)

1 sin 1 sin 2 1 sin 1 sin

1 sin 1 sin

x x x x

x x

2

2

2 1 1 sin

2sin

1 cos

sin

2cos2

2sin cos2 2

x

x

x

x

x

x x

cot2

x

1

1

1 sin 1 sin. . . cot

1 sin 1 sin

cot cot . . .2 2

x xL H S

x x

x xR H S

Question 11:

Prove 1 11 1 1 1tan cos , 1

4 21 1 2

x xx x

x x

: put cos 2x Hint

Solution 11:

Let, cos 2x then, 11

cos .2

x

Thus, we have:



1 1 1. . . tan

1 1

x xL H S

x x

1 1 cos 2 1 cos 2tan

1 cos2 1 cos 2

2 2

1

2 2

2cos 2 2sintan

2cos 2 2sin

2 cos 2 sin

tan2 cos 2 sin

1

1

cos sintan

cos sin

1 tantan

1 tan

1 1 1 1 1tan 1 tan tan tan tan tan1

x yx y

xy

11cos R.H.S.

4 4 2x

Question 12:

Prove 1 19 9 1 9 2 2sin sin

8 4 3 4 3

Solution 12:

19 9 1L.H.S. = sin

8 4 3

19 1sin

4 2 3

19 1cos

4 3

……(1) 1 1sin cos

2x x

Now, let 1 1cos

3x



Then, 1

cos3

x

21 2 2

sin 1 .3 3

x

1 1 12 2 1 2 2sin cos sin

3 3 3x

19 2 2L.H.S. = sin R.H.S.

4 3

Question 13:

Solve 1 12tan cos tan 2cosecx x

Solution 13:

1 12tan cos tan 2cosecx x

1 1 1 1

2 2

2cos 2tan tan 2cosec 2 tan tan

1 cos 1

x xx x

x x

2

2cos2cosec

1 cos

xx

x

2

2cos 2

sin sin

x

x x

cos sinx x

tan 1x

4x

Question 14:

Solve 1 11 1tan tan , 0

1 2

xx x

x

Solution 14:

1 11 1tan tan

1 2

xx

x



1 1 1 1 1 11tan 1 tan tan tan tan tan

2 1

x yx x x y

xy

13tan

4 2x

1tan6

x

tan6

x

1

3x

Question 15:

Solve 1sin tan , 1x x is equal to

(A) 21

x

x (B)

2

1

1 x

(C) 2

1

1 x (D)

21

x

x

Solution 15:

2tan sin

1

xy x y

x

Let 1tan .x y Then,

1 1 1

2 2sin tan sin

1 1

x xy x

x x

1 1

2 2sin tan sin sin

1 1

x xx

x x

The correct answer is D.



Question 16:

Solve 1 1sin 1 2sin2

x x , then x is equal to

(A) 1

0,2

(B) 1

1,2

(C) 0 (D) 1

2

Solution 16:

1 1

1 1

1 1

1 2

1 2

1 1 2

sin 1 2sin2

2sin sin 12

2sin cos 1 ...(1)

Let sin sin cos 1 .

cos 1

sin cos 1

x x

x x

x x

x x x

x

x x

Therefore, from equation (1), we have

1 2 12cos 1 cos 1x x

Let, sin .x y Then, we have:

1 2 12cos 1 sin cos 1 siny y

1 12cos cos cos 1 siny y

12 cos 1 siny y

1 sin cos 2 cos2y y y

21 sin 1 2siny y 22sin sin 0y y

sin 2sin 1 0y y

1sin 0

2y or

10

2x OR x

1When , it can be observed that :

2x



1 11 1L.H.S. sin 1 sin

2 2

1 11 1sin 2sin

2 2

1 1sin

2

. . .6 2

R H S

1

2x is not a solution of the given equation.

Thus, 0.x

Hence, the correct answer is C.

Question 17:

Solve 1 1tan tan

x x y

y x y

is equal to

(A) 2

(B)

3

(C) 4

(D)

3

4

Solution 17:

1 1tan tan

x x y

y x y

1 1 1 1tan tan tan tan1

1

x x y

x yy x yy y

xyx x y

y x y

1tan

x x y y x y

y x y

y x y x x y

y x y

2 21

2 2tan

x xy xy y

xy y x xy



2 21 1

2 2tan tan 1

4

x y

x y

Hence, the correct answer is .C


Class XII NCERT Maths Chapter 2- Inverse …...Class XII – NCERT – Maths Chapter 2- Inverse...

Documents

Transcript of Class XII NCERT Maths Chapter 2- Inverse …...Class XII – NCERT – Maths Chapter 2- Inverse...